# Spin cobordism v.s. KO theory in low or in any dimensions

It seems that from this webpage, the spin cobordism is equivalent to KO theory in low dimension.

If we denote the $$p$$-torsion part (mean $$\mathbb{Z}_{p^n}$$ for some $$n$$) $$\Omega_d(BG)_p.$$

Question 1: Then do we have $$\Omega_d^{spin}(BG)_p = ko_d(BG)_p?$$ for $$p=2$$ and free part, for $$d\le 7$$? (how about higher $$d>7$$?)

And $$\Omega_d^{spin}(BG)_p = \Omega_d^{SO}(BG)?$$ for $$p \neq 2$$ and $$p$$ is an odd prime?

Namely, the 2-torsion and free part of $$Mspin$$ and $$KO$$ is the same. If there is an odd $$p$$ torsion, we need to consider localization at odd prime by $$MSO$$ cohomology. Is this correct?

Question 2: If this is a statement about the spectra, not just about stable homotopy groups, and thus within these spin cobordism and ko theory, do they completely coincide for any dimensions $$d$$, instead of just $$d \leq 7$$?

• i) By ko, do you mean the connective real K-theory? ii) What is G here? A finite group? Discrete greoup? Compact Lie group? Any topological group? – user43326 Oct 16 '18 at 4:54
• @user43326 in the context wonderich is asking about, $\mathit{ko}$ is indeed connective real $K$-theory, and I believe $G$ can be any compact Lie group. – Arun Debray Oct 22 '18 at 4:34

First of all, let me say that the page you are quoting is a little bit misleading if not inaccurate on the Anderson-Brown-Peterson splitting.

$$ko\langle 4n(J)\rangle$$ should read $$\Sigma ^{4n(J)} ko$$ and $$ko\langle 4n(J)-2\rangle$$ should read $$\Sigma ^{4n(J)-4} ko\langle 2\rangle$$

With this correction, at 2, $$Mspin \wedge BG$$ splits whose bottom piece is $$ko\wedge BG$$, other pieces are at least 7-connected since the "next bottom" piese is $$\Sigma ^8ko \wedge BG$$ as is pointed out by ArunDebray, corresponding to the partition $$J=(2)$$.

Thus the answer to your questions, at the prime 2 is that

1. We have an isomorphism up to $$d\leq 7$$
2. The map is always surjective, but the kernel is in general non-trivial for $$d\geq 8$$.
• Are you sure it's a $\Sigma^4\mathit{ko}$, and not a $\Sigma^8\mathit{ko}$? $\pi_5\mathit{MSpin}$ and $\pi_6\mathit{MSpin}$ both vanish, but if $\mathit{MSpin}$ had a $\Sigma^4\mathit{ko}$ summand, they would both contain a $\mathbb Z/2$ summand. – Arun Debray Oct 22 '18 at 16:51
• @ArunDebray You are right, $n(J)$ even, so we get $\Sigma ^8ko$. With $n(J)$ odd sequences, we are not allowed to have 1 so the lowest is $\Sigma 8ko<2>$. Presumably in the range the op asks, there is no HZ/2 summand, I will correct my answer later, thanks. – user43326 Oct 23 '18 at 8:31
• Thanks for the answer/comments - I hope it is correct. +1 – wonderich Nov 20 '18 at 0:24
• I asked one more related question for the Pin cases --- please feel free to answer/comments - thanks - I am not sure what will be the related K theory. – wonderich Nov 20 '18 at 0:36